Given: \(\lambda=2\), and \(x=5\).
Using the Poisson distribution formula:
\(P(X=x)=\left(e^{-\lambda} \lambda^{x}\right) / x !\)
\(P(X=5)=\left(e^{-2} 2^{5}\right) / 5 !\)
\(P(X=6)=0.036\)
Answer: The probability of arrival of 5 customers per minute is \(3.6 \%\).